Let's solve each part **step by step** based on the provided equations and data. ### **Given:** - **Activity coefficients:** \[ \ln \gamma_1 = A x_2^2 \] \[ \ln \gamma_2 = A x_1^2 \] \[ A = 2.777 - 0.005237T \] - **Vapor pressure equations (Antoine):** \[ \ln P_1^* = 16.59158 - \frac{3638.27}{T} - 33.424 \] \[ \ln P_2^* = 14.25326 - \frac{2665.54}{T} - 53.424 \] Where: - \( T \) in K - \( P_1^*, P_2^* \) in kPa --- ## **(a) γ₁ and γ₂ for T = 318.15 K and x₁ = 0.25** ### **Step 1: Find A at T = 318.15 K** \[ A = 2.777 - 0.005237 \times 318.15 \] \[ A = 2.777 - 1.666 = 1.111 \] --- ### **Step 2: Find γ₁ and γ₂** \[ x_1 = 0.25,\quad x_2 = 1 - 0.25 = 0.75 \] \[ \ln \gamma_1 = A x_2^2 = 1.111 \times (0.75)^2 = 1.111 \times 0.5625 = 0.625 \] \[ \gamma_1 = \exp(0.625) = 1.868 \] \[ \ln \gamma_2 = A x_1^2 = 1.111 \times (0.25)^2 = 1.111 \times 0.0625 = 0.0694 \] \[ \gamma_2 = \exp(0.0694) = 1.072 \] --- ### **Final Answers (a):** \[ \boxed{ \gamma_1 = 1.87,\quad \gamma_2 = 1.07 } \] --- ## **(b) γ₁ and γ₂ for T = 318.15 K and x₁ = 0.85** \[ x_1 = 0.85,\quad x_2 = 0.15 \] \[ \ln \gamma_1 = 1.111 \times (0.15)^2 = 1.111 \times 0.0225 = 0.025 \] \[ \gamma_1 = \exp(0.025) = 1.025 \] \[ \ln \gamma_2 = 1.111 \times (0.85)^2 = 1.111 \times 0.7225 = 0.803 \] \[ \gamma_2 = \exp(0.803) = 2.23 \] --- ### **Final Answers (b):** \[ \boxed{ \gamma_1 = 1.03,\quad \gamma_2 = 2.23 } \] --- ## **(c) γ₁ and γ₂ for P = 101.3 kPa, x₁ = 0.40** Still at \( T = 318.15 \) K, so \( A = 1.111 \): \[ x_1 = 0.40,\quad x_2 = 0.60 \] \[ \ln \gamma_1 = 1.111 \times (0.60)^2 = 1.111 \times 0.36 = 0.400 \] \[ \gamma_1 = \exp(0.400) = 1.49 \] \[ \ln \gamma_2 = 1.111 \times (0.40)^2 = 1.111 \times 0.16 = 0.178 \] \[ \gamma_2 = \exp(0.178) = 1.19 \] --- ### **Final Answers (c):** \[ \boxed{ \gamma_1 = 1.49,\quad \gamma_2 = 1.19 } \] --- ## **(e) The azeotropic pressure and composition for T = 318.15 K** ### **Step 1: Find \(P_1^*\) and \(P_2^*\) at \(T = 318.15\) K** #### **For \(P_1^*\):** \[ \ln P_1^* = 16.59158 - \frac{3638.27}{318.15} - 33.424 \] \[ \frac{3638.27}{318.15} = 11.44 \] \[ \ln P_1^* = 16.59158 - 11.44 - 33.424 = -28.272 \] \[ P_1^* = \exp(-28.272) \approx 5.19 \times 10^{-13}\ \text{kPa} \] This value is unreasonably low, indicating a possible typo in the Antoine equation constants. Let's check \(P_2^*\): #### **For \(P_2^*\):** \[ \ln P_2^* = 14.25326 - \frac{2665.54}{318.15} - 53.424 \] \[ \frac{2665.54}{318.15} = 8.38 \] \[ \ln P_2^* = 14.25326 - 8.38 - 53.424 = -47.55074 \] \[ P_2^* = \exp(-47.55) \approx 2.34 \times 10^{-21}\ \text{kPa} \] Again, this is extremely low. These values are not physically meaningful (likely due to incorrect constants or missing parentheses in the equations). Usually, the Antoine equation is: \[ \ln P^* = A - \frac{B}{T + C} \] But given the problem, let's proceed with the azeotrope calculation in principle. --- ### **Step 2: Azeotrope Condition** At an azeotrope: \[ y_1 = x_1, \quad y_2 = x_2 \] So, \[ P = x_1 \gamma_1 P_1^* + x_2 \gamma_2 P_2^* \] Also, at the azeotrope: \[ \frac{y_1}{y_2} = \frac{x_1}{x_2} = \frac{\gamma_1 P_1^*}{\gamma_2 P_2^*} \] \[ \frac{x_1}{1 - x_1} = \frac{\gamma_1 P_1^*}{\gamma_2 P_2^*} \] Take logarithms: \[ \ln \left(\frac{x_1}{1 - x_1}\right) = \ln \gamma_1 + \ln P_1^* - \ln \gamma_2 - \ln P_2^* \] But using: \[ \ln \gamma_1 = A (1-x_1)^2 \] \[ \ln \gamma_2 = A (x_1)^2 \] So, \[ \ln \left(\frac{x_1}{1 - x_1}\right) = A (1-x_1)^2 - A (x_1)^2 + \ln P_1^* - \ln P_2^* \] \[ \ln \left(\frac{x_1}{1 - x_1}\right) = A [(1-x_1)^2 - x_1^2] + \ln \left(\frac{P_1^*}{P_2^*}\right) \] Now, plug values for \(A\) and (if the vapor pressures were correct) \(P_1^*/P_2^*\). #### **Given the extremely low vapor pressures, it's best to use the procedure above once correct constants are obtained.** --- ## **Summary of Answers** ### (a) \[ \boxed{\gamma_1 = 1.87,\ \gamma_2 = 1.07} \] ### (b) \[ \boxed{\gamma_1 = 1.03,\ \gamma_2 = 2.23} \] ### (c) \[ \boxed{\gamma_1 = 1.49,\ \gamma_2 = 1.19} \] ### (e) - **Azeotrope composition and pressure:** Use the equation above with correct \(P^*\) values. --- **If you have correct Antoine constants or tabulated vapor pressures for methanol and ethanol at 318.15 K, plug them in the equation for the azeotrope calculation. Otherwise, the procedure above is correct for solving the azeotrope; just input accurate \(P_1^*\) and \(P_2^*\) values.** --- **If you want to continue with the azeotrope calculation using estimated vapor pressures, let me know!**

Certainly! Based on the provided data and equations, I will now generate the answer for part **(d)**. --- ## **(d) Calculate the vapor-liquid equilibrium (VLE) at \( T = 318.15\,K \) and \( P = 101.3\,kPa \). Find the equilibrium compositions \( x_1 \) and \( y_1 \).** --- ### **Step 1: Determine the vapor pressures \( P_1^* \) and \( P_2^* \) at \( T=318.15\,K \).** Given Antoine equations: \[ \ln P_1^* = 16.59158 - \frac{3638.27}{T} - 33.424 \] \[ \ln P_2^* = 14.25326 - \frac{2665.54}{T} - 53.424 \] Calculations: \[ \frac{3638.27}{318.15} \approx 11.44 \] \[ \ln P_1^* = 16.59158 - 11.44 - 33.424 = -28.272 \] \[ P_1^* = e^{-28.272} \approx 5.19 \times 10^{-13}\, \text{kPa} \] Similarly, \[ \frac{2665.54}{318.15} \approx 8.38 \] \[ \ln P_2^* = 14.25326 - 8.38 - 53.424 = -47.5507 \] \[ P_2^* = e^{-47.5507} \approx 1.19 \times 10^{-21}\, \text{kPa} \] **Note:** These extremely low vapor pressure values suggest potential errors or typo in the constants, but we'll proceed with the methodology. --- ### **Step 2: Write the Raoult's law for each component at equilibrium:** \[ P y_1 = x_1 \gamma_1 P_1^* \] \[ P y_2 = x_2 \gamma_2 P_2^* \] Since \( y_1 + y_2 = 1 \), at equilibrium: \[ \frac{y_1}{y_2} = \frac{\gamma_1 P_1^* x_1}{\gamma_2 P_2^* x_2} \] And, \[ \boxed{ \frac{y_1}{y_2} = \frac{P y_1}{P y_2} = \frac{\text{partial pressures}} } \] Given the vapor pressure data is not physically meaningful here, but the general relation for vapor-liquid equilibrium is: \[ y_1 = \frac{\gamma_1 P_1^* x_1}{P} \] \[ y_2 = \frac{\gamma_2 P_2^* x_2}{P} \] and \[ \text{At equilibrium:} \quad y_1 = \frac{\gamma_1 P_1^* x_1}{P} \] Similarly for \( y_2 \). --- ### **Step 3: Find the equilibrium compositions \( x_1 \) and \( y_1 \).** Assuming ideality and neglecting activity coefficients for this estimation (since \( \gamma_1, \gamma_2 \) are close to 1 in previous parts), the relation simplifies to: \[ y_1 = \frac{P_1^* x_1}{P} \] \[ y_2 = \frac{P_2^* x_2}{P} \] Given the vapor pressure constants lead to negligible vapor pressures, we can interpret that the mixture is not vaporizing significantly at this temperature, or that the vapor pressure data needs correction for practical purposes. --- ### **Final Note:** Due to the unphysical vapor pressure values obtained, exact numerical solutions for \( x_1 \) and \( y_1 \) at \( P=101.3\,kPa \) cannot be reliably calculated from the data provided. **However, the general approach is:** 1. Calculate \( P_1^* \) and \( P_2^* \) at the temperature. 2. Use Raoult’s law with activity coefficients to relate \( x_1, y_1 \). 3. Solve for \( y_1 \) and \( x_1 \) at the given total pressure \( P \). --- ## **Summary:** - **The key relations:** \[ y_1 = \frac{\gamma_1 P_1^* x_1}{P} \] \[ y_2 = \frac{\gamma_2 P_2^* x_2}{P} \] - **The sum of vapor phase mole fractions:** \[ y_1 + y_2 = 1 \] - **Thus, to find \( x_1 \):** \[ x_1 = \frac{P y_1}{\gamma_1 P_1^*} \] and similarly for \( x_2 \). --- **In conclusion:** - **Without accurate vapor pressure data, a precise numerical answer cannot be provided here.** - **The methodology above outlines how to determine vapor-liquid equilibrium compositions at given temperature and pressure.** --- Please provide corrected vapor pressure data if available for a more accurate calculation!